Boost converter

The mode of operation and the change of the input resistance by the duty cycle was explained in the boost converter basics chapter. For the solar module optimiser it is important that as wide a range of the input resistance as possible is covered. However, for the basis of the calculation of the elements of the boost converter, a limited range shall be used. The minimum input voltage $U_{in}$ as well as the maximum output voltage $U_{out}$ are obtained from the specified characteristics. The maximum duty cycle $d_{max}$ is calculated according to equation \eqref{eq:dutymax}.

\[\label{eq:dutymax} d_{max} = 1 - \frac{V_{in}}{V_{out}} = 1 - \frac{15V}{50V} = 0.7\]

Then, equation \eqref{eq:dutymin} can be used to determine the minimum duty cycle.

\[\label{eq:dutymin} d_{min} = 1 - \left(\frac{V_{in}}{V_{out}}\right) \cdot (1-d_{max}) = 1 - \left(\frac{30V}{15V}\right) \cdot (1-0,7) = 0,4\]

In order to take into account the different input voltages $V_{in}$, the calculation of the inductance $L$ is performed with the minimum duty cycle $d_{min}$ and the maximum input voltage $V_{max}$. The size of the current ripple $\Delta i_{L}$ is a design decision and is chosen to be $25\%$ of the maximum input current $I_{in}$. The switching frequency $f_{s}$ is set to $50kHz$ to keep the size of the inductance small and to choose a range that is not perceptible to the human ear.

\[\label{eq:induc} L = \frac{V_{max}d_{min}}{\Delta i_{L} f_{s}} = \frac{30V \cdot 0,4}{2,5A \cdot 50 kHz} = 96 \mu H\]

The equation \eqref{eq:induc} determines the minimum required inductance $L$ for the boost converter. A toroidal choke with an inductance of $L=100 \mu H$ and a maximum current carrying capacity of $I_{L}=10A$ is selected.

To ensure a uniform output voltage, the output voltage delta $\Delta U_{out}$ shall be $0.1V$ maximum and the output current $I_{out}$ is fixed at $2A$. The capacitor $C$ on the output side is now calculated according to \eqref{eq:Cout}.

\[\label{eq:Cout} C_{out} = \frac{I_{out}D_{max}}{\Delta V_{out} f_{s}} = \frac{2A \cdot 0,7}{0,1V \cdot 50 kHz} = 280 \mu F\]

The capacitance of the output capacitor should be greater than the calculated value and is thus chosen to be $C_{out} = 470 \mu F$. The voltage rating is chosen to be $V_{C_{outmax}}=68V$, as the maximum output voltage is $V_{out_{max}}=50V$.